1. STEPS: x4 + 3x3 – 12x2 + 12x < 0 x(x – 2)2 (x + 3) < 0
Graph the polynomial function and then find the intervals that apply to the inequality.
STEPS:
Find and graph zeros. Find intervals.
Cross vs. Bump
Find direction of endpoints: + up, – down (on right) and turning points.
x4 + 3x3 – 12x2 + 12x < 0
In factored form.
x(x – 2)2 (x + 3) < 0

2. STEP 1: Find and graph zeros. Find intervals.
x(x – 2)2(x + 3) < 0
x = 0, 2, –3
Intervals:
(-, -3) (-3, 0) (0, 2) (2, )

3. STEP 2: Cross vs. Bump x(x – 2)2(x + 3) < 0
Even exponents are always positive so they indicate a bump of the x-axis. Odd exponents can be negative so they cross the x-axis.
BUMP

4. STEP 3: Find endpoint directions: + up, – down (on right) and turning points to sketch the graph.
x(x – 2)2(x + 3) < 0
If not given the Leading Coefficient would be: x·x2·x = x4
POSITIVE leading coefficient means up on right and an EVEN exponent means same direction on left. x4 indicates 3 turning points.
3 turning points

5. There’s no need for exact points for peaks/valleys because we’re looking for specific intervals below the x-axis (< 0).
x(x – 2)2(x + 3) < 0
(1, 4)
Zeros: 0, 2, –3
Intervals: (-, -3) (-3, 0) (0, 2) (2, )
(-3, 0)
NOTE: If it were < the interval would be [-3, 0] and {2}.

6. Since we’re just looking for intervals, polynomial inequalities can be graphed on a number line.
x(x – 2)2(x + 3) < 0
Zeros: 0, 2, –3
Intervals: (-, -3) (-3, 0) (0, 2) (2, )
(-3, 0) or -3 < x < 0